A committee consisting of at least three members is to be formed from a group
of six boys and six girls such that it always has a boy and a girl. Find the
number of ways to form such committee.
First we count all committees of any size without regard to size or sex.
(This will even include a committee of size 0.) Then we'll subtract the
number of unacceptable committees.
In choosing a committee without regard to size or sex, we
tell each of the 12 people one of two things:
1. YES, you will serve on the committee
2. NO, you will NOT serve on the committee.
If there were no restrictions with regard to size or sex the answer
would be 212.
Now we subtract the unacceptable committees.
We may not tell all the girls NO, for that would leave an all-boy committee.
So that is the number of ways we could tell all the girls NO, and tell the 6
boys YES or NO, which would be 26.
We also may not tell all the boys NO, for that would leave an all-girl
committee. So that is the number of ways we could tell all the boys NO, and
tell the 6 girls YES or NO, which would also be 26.
That would be 2·26 or 27 to subtract. However both those
have 1 case in common, the case where all 12 people are told NO, so we must
avoid subtracting that case twice, so we can only subtract 1 less than that or
27 - 1
So without regard to the committee size, there are
212 - (27 - 1) or 212 - 27 + 1 committees.
Now we only have to subtract the number of committees of size 2 consisting
of 1 boy and 1 girl. There are 6 ways to choose the girl and 6 ways to
choose the boy. That's 6·6 or 36 committees of 2 to subtract.
[Note: We have already subtracted the committees of sizes 0 and 1. The
committee of size 0 was the one we had to avoid counting twice above, and a
committee of 1 cannot contain both sexes.]
Final answer: 212 - 27 + 1 - 36 = 4096-128+1-36 = 3933
Edwin
